Talk:Clarkkkkson
Originally defined Clarkkkkson is actually between \( 10 \uparrow\uparrow\uparrow 2 \) and \( 10 \uparrow\uparrow\uparrow 3 \), as weak operators can't provide even pentational growth. Ikosarakt1 (talk) 21:01, January 7, 2013 (UTC) :It's sort of a salad number FB100Z • talk • 22:38, January 7, 2013 (UTC) Wait, no. I reconsider that and conclude that it grows pentationally however, and actual value is between tritri and g1. I am interested for more tight bounds, despide that it's just "salad" number. Ikosarakt1 (talk) 13:47, January 14, 2013 (UTC) Some weak upper bounds and stuff: * \(n! < n^n\) * \(n!! < n^n \cdot (n - 1)^{n - 1} \cdot (n - 2)^{n - 2} \ldots = (n^n)^n = n^{n^2}\) * \(n!!! < \left(n^{n^2}\right)^n = n^{n^3}\) * \(n!c < n^{n^c}\) * \(\text{hypf}(c, 2, n) < n^{n^c} \cdot (n - 1)^{(n - 1)^c} \ldots < n^{n \cdot n^c} = n^{n^c} \uparrow n\) * \(\text{hypf}(c, 3, n) < n^{n^c} \uparrow (n - 1)^{(n - 1)^c} \uparrow \ldots < n^{n^c} \uparrow\uparrow n\) * \(\text{hypf}(c, p, n) < n^{n^c} \uparrow^{p - 2} (n - 1)^{(n - 1)^c} \uparrow^{p - 2} \ldots < n^{n^c} \uparrow^{p - 1} n < n^{n^c} \uparrow^{p - 1} n^{n^c}\) * \(\text{ck}(c, p, n, r) < n^{n^c} \uparrow^p 2^r\) ** \(f(a) = a \uparrow^c a \implies f^n(a) \leq a \uparrow^{c + 1} 2^n\). I'm too lazy to prove this, but it "looks" right. * \(\text{ck}(n, n, n, n) < n^{n^n} \uparrow^n 2^n\) * \(n^{n^n} \uparrow^n 2^n < n \uparrow^{n + 1} n\) (rather sketchy comparison) * \(\text{ck}^{n}(n) < n \uparrow^{n + 2} 2^n\) * ¥ < {K, , K + 2} The variables have to be sufficiently large, but on order of the lynz the bounds will definitely work. I'm convinced that ¥ << Graham's number. Also, I assumed the strong hyper-operators instead of the weak ones, so these bounds are not very tight. Good enough for government work, as they say. FB100Z • talk • 00:14, January 15, 2013 (UTC) Hope y'all don't mind me using this page as scratch paper. * \(a ↓↓ b = a^{a^{b - 1}} < a^{a^a} = a ↑↑ 3\) for \(b - 1 < a\) * \(a ↓↓↓ 2 = a ↓↓ a < a ↑↑ 3\); note that \(a - 1 < a\) * \(a ↓↓↓ 3 = a ↓↓ a ↓↓ a < (a ↑↑ 3) ↑↑ 3 < a ↑↑ 3^2\) * \(a ↓↓↓ 4 = a ↓↓ a ↓↓ a ↓↓ a < ((a ↑↑ 3) ↑↑ 3) ↑↑ 3 < a ↑↑ 3^3\) * \(a ↓↓↓ 5 = a ↓↓ a ↓↓ a ↓↓ a ↓↓ a < (((a ↑↑ 3) ↑↑ 3) ↑↑ 3) ↑↑ 3 < a ↑↑ 3^4\) * \(a ↓↓↓ b < a ↑↑ 3^{b - 1}\) * \(a ↓↓↓↓ 2 = a ↓↓↓ a < a ↑↑ 3^{a - 1}\) * \(a ↓↓↓↓ 3 = a ↓↓↓ a ↓↓↓ a < (a ↑↑ 3^{a - 1}) ↑↑ 3^{a - 1} < a ↑↑ (3^{a - 1})^2\) * \(a ↓↓↓↓ 4 = a ↓↓↓ a ↓↓↓ a ↓↓↓ a < ((a ↑↑ 3^{a - 1}) ↑↑ 3^{a - 1}) ↑↑ 3^{a - 1} < a ↑↑ (3^{a - 1})^3\) * \(a ↓↓↓↓ b < a ↑↑ 3^{(a - 1)(b - 1)}\) * \(a ↓↓↓↓↓ b < a ↑↑ 3^{(a - 1)^2(b - 1)}\) * \(a ↓↓↓↓↓↓ b < a ↑↑ 3^{(a - 1)^3(b - 1)}\) * \(a ↓^{n + 3} b < a ↑↑ 3^{(a - 1)^n(b - 1)}\) * \(\text{hypf}(c, 2, n) < n^{n^c} ↓ n = n^{n^c} ↑ n\) * \(\text{hypf}(c, 3, n) < n^{n^c} ↓↓ n < n^{n^c} ↑↑ 3\) * \(\text{hypf}(c, 4, n) < n^{n^c} ↓↓↓ n < n^{n^c} ↑↑ 3^{n - 1}\) * \(\text{hypf}(c, 5, n) < n^{n^c} ↓↓↓↓ n < n^{n^c} ↑↑ 3^{(n^{n^c} - 1)(n - 1)}\) * \(\text{hypf}(c, p, n) < n^{n^c} ↓^{p - 1} n < n^{n^c} ↑↑ 3^{(n^{n^c} - 1)^{p - 4}(n - 1)}\) ** This is less than \(n ↑↑↑ 3\) for sufficiently large inputs but I can't prove it * \(\text{ck}(c, p, n, r) < n ↑↑↑ 3^{r - 1}\) * \(\text{ck}(n) < n ↑↑↑ 3^{n - 1} < n ↑↑↑↑ 3\) * \(\text{ck}^n(n) < n ↑↑↑↑ 3^{n - 1}\) * ¥ < {K, , 4} which is compatible with Ikosarakt's guess. If only the inventor of ¥ had used the strong operators. FB100Z • talk • 19:44, January 15, 2013 (UTC) Where \(f(x) = x \uparrow^c a\), it turns out that \(f^n(x)\) is bounded above by \(x \uparrow^c (a + n)\). Clearly I have overestimated the down arrows. FB100Z • talk • 00:10, January 17, 2013 (UTC) We can do better for weak hyper-operators: \((a \lbrace c \rbrace b)\lbrace c \rbrace d < (a \lbrace c \rbrace (b+d)\), using left associative polyates lemma, here \(\lbrace c \rbrace\) is any operator. Thus: \(a \downarrow\downarrow a < a \uparrow\uparrow 3\) \(a \downarrow\downarrow\downarrow a < a \uparrow\uparrow 3a\) \(a \downarrow_{c} a < a \uparrow\uparrow 3a^{c-2}\) Ikosarakt1 (talk) 18:26, January 18, 2013 (UTC) :It even looks like \((a ↑^c b) ↑^c d \approx a ↑^c (b + 1)\) (yielding the bound \(a \uparrow^c (b + 2)\)). :For example \((a ↑↑ 3) ↑↑ 2 = (a^{a^a})^{(a^{a^a})} = a^{a^a \cdot a^{a^a}} = a^{a^{a + a^a}} \approx a^{a^{a^a}}\). FB100Z • talk • 23:04, January 18, 2013 (UTC) Arbitrary section break I'm now totally convinced that \(f(x) = x \uparrow^c a\) implies \(f^n(x) < x \uparrow^c (a + n)\). Since I'm a googologist and I name everything, I will call this the "arrow iteration lemma." * \(a ↓↓ b < a ↑↑ 3\) if \(b - 1 < a\) * \(a ↓↓↓ 2 = a ↓↓ a < a ↑↑ 3\) * \(a ↓↓↓ 3 = a ↓↓ a ↓↓ a < (a ↑↑ 3) ↑↑ 3 < a ↑↑ 5\) by the arrow iteration lemma * \(a ↓↓↓ 4 = a ↓↓ a ↓↓ a ↓↓ a < ((a ↑↑ 3) ↑↑ 3) ↑↑ 3 < a ↑↑ 6\) * \(a ↓↓↓ b < a ↑↑ (b - 2)\) * \(a ↓↓↓↓ 2 = a ↓↓↓ a < a ↑↑ (a - 2)\) * \(a ↓↓↓↓ 3 = a ↓↓↓ a ↓↓↓ a < (a ↑↑ (a - 2)) ↑↑ (a - 2) < a ↑↑ a\) * \(a ↓↓↓↓ 4 = a ↓↓↓ a ↓↓↓ a ↓↓↓ a < ((a ↑↑ (a - 2)) ↑↑ (a - 2)) ↑↑ (a - 2) < a ↑↑ (a + 1)\) * \(a ↓↓↓↓ b < a ↑↑ (a + b - 2)\) * \(a ↓↓↓↓↓ 2 < a ↑↑ (2a - 2)\) * \(a ↓↓↓↓↓ b < a ↑↑ (2a + b - 2)\) * \(a ↓↓↓↓↓↓ b < a ↑↑ (3a + b - 2)\) * \(a ↓^{n + 3} b < a ↑↑ (na + b - 2)\) FB100Z • talk • 23:26, January 18, 2013 (UTC) I tried to trace \(f(n) = 3 \downarrow_{n} 3\) and your bounds fail at n=5. Particularly: \(3 \downarrow 3 = 27\) \(3 \downarrow\downarrow 3 = 3^{3^2} = 19683\) \(3 \downarrow\downarrow\downarrow 3 = (3 \downarrow\downarrow 3) \downarrow\downarrow 3 = 19683 \downarrow\downarrow 3 = 19683^{19683^{19682}} = 3^{3^{20}}\) (weak "tritri") \(3 \downarrow\downarrow\downarrow\downarrow 3 = (3 \downarrow\downarrow\downarrow 3) \downarrow\downarrow\downarrow 3 = 3^{3^{20}} \downarrow\downarrow\downarrow 3\) \(n \downarrow\downarrow\downarrow 2 > n \uparrow\uparrow 2 = n^n\) \(n \downarrow\downarrow\downarrow 3 > n \uparrow\uparrow 3 = n^{n^n}\) \(n \downarrow\downarrow\downarrow n > n \uparrow\uparrow n\) \(3^{3^{20}} \downarrow\downarrow\downarrow 3\) > \(3^{3^{20}} \uparrow\uparrow 3 > 3 \uparrow\uparrow 4 = N\) \(3 \downarrow\downarrow\downarrow\downarrow\downarrow 3 = (3 \downarrow\downarrow\downarrow\downarrow 3) \downarrow\downarrow\downarrow\downarrow 3 > N \downarrow\downarrow\downarrow\downarrow 3 = (N \downarrow\downarrow\downarrow N) \downarrow\downarrow\downarrow N\) \(N \downarrow\downarrow\downarrow N > N \uparrow\uparrow N > 3 \uparrow\uparrow N > 3 \uparrow\uparrow (3 \uparrow\uparrow 4)\) But by last your bounds: \(3 \downarrow_{5} 3 < 3 \uparrow\uparrow 7\). Something here does not converge. Ikosarakt1 (talk) 12:43, January 21, 2013 (UTC) I see there was been a lot of talk about the weak operators, especially in relation to the clarkkkkson. There seems to be a lot of confusion about what these operators are doing in relation to the strong-operators. I have known for quite some time however that for any strong-operator there is a weak-operator of roughly twice the order with roughly the same strength. I meant to write an article about it eventually but I never got around to writing it. Although it seems counter intuitive it isn't too difficult to verify. There seems to be an assumption that the weak-operators never make it past tetration. If < means "grows slower that" we can say: bvvp < b^^p since bvvp = b^b^(p-1) What happens however when we recursively plug bvvp into itself and resolve from left to right? We have: bvvvp = bvvbvvbvvbvv ... vvbvvb = b^b^(b-1) vvbvvb ... bvvb At this point we need to solve b^b^(b-1)vvb where b^b^(b-1) is the base. we would obtain: bvvbvvb = (b^b^(b-1))^(b^b^(b-1))^b-1. Now we can imediately see, that this must be larger than b^b^b^(b-1). In fact we can see that Xvvp > X^X. We know that iterating X^X grows as fast as tetration from experience with steinhaus triangles, so we conclude: bvvvp = b^^p Given this, we can approximate bvvvvp as roughly equivalent to solving a series of tetrations from left to right. thus bvvvvp ~ b^^b^^ ... ^^b^^b Here the Left Associatives Polyate Lemma (LAPL) ''is vital. I originally devised this lemma as part of a proof for ''Moser ''< ''Graham's No. ''but it has a few applications. The lemma basically states that: (b{c}n){c}m ~< b{c}(n+m) I was also suppose to write an article on this lemma and never got around to it. Using it however we obtain that b^^b^^ ... ^^b^^b w/p bs ~ b^^(bp) So we have the result that: b vvvv p ~ b^^(bp) in otherwords, it's "slightly" stronger than b^^p. Now bvvvvvp or bv(5)p makes a dramatic jump since: bv(5)p = bvvvvbvvvv ... vvvvbvvvvb ~ b^^(bp)vvvvbvvvv ... vvvvbvvvvb ~ (b^^bp)^^(p(b^^bp))vvvv ... vvvvb we can see that (b^^bp)^^(p(b^^bp)) > b^^b^^(bp), so again we get a case where iterating bvvvvp and solving from left to right results in something approximating a hyper-operator. Thus: bv(5)p = b^^^p At this point it should be clear enough that if we continue using LAPL we would be able to show eventually that: bv(k)p = b^((k+1)/2)p ie. bv(3)p = b^(2)p bv(5)p = b^(3)p bv(7)p = b^(4)p bv(9)p = b^(5)p Admittedly this proof is informal, but it should at least make it clear how a rigorous proof could be built, especially since my LAPL also includes a lowerbound of b{c}(n+m-1). That wouldn't be enough to fundamentally change the conclusion. What all this implies in relation to the ''clarkkkkson ''is that it should be on par with numbers like ''Graham's No. ''In fact, I remember that I once estimated that it was much larger, but would still never get very far none the less (since it grows over time, but ''Lynz ''grows very slow relative to the function K(n)). Sbiis Saibian (talk) 17:41, January 21, 2013 (UTC) I initially thought to use your LAPL to show that \(a \downarrow_{c} b < a \uparrow\uparrow (3*b^{c-2})\), and consider \(f(n) = 3 \downarrow_{n} 3\), but this bound fails me at n=5. Ikosarakt1 (talk) 13:15, January 22, 2013 (UTC) Where did you first here mention of LAPL? I was working on an article for it, but never finished it. I did however briefly make mention of the lemma in a community email (I periodically send out group-emails to all the "big names" in googology like Robert Munafo, Jonathan Bowers, Chris Bird, Giga Gerard of the Big Psi project to name a few). I'm not sure if anyone from the ''googology wiki ''recieves these emails. In anycase, LAPL does not actually make mention of the weak-operators. Rather it is an important bounding lemma for applying any of the hyper-operators in a left to right order. It can be used to both show that certain functions grow just as fast as the hyper-operators, and it can also be used to show that certain constructions are vastly smaller. Actually the first occurance of the lemma is when I was trying to prove gaggol^^gaggol < greagol. Here I found that not only was gaggol^^gaggol < 10^^(gaggol*gaggol), but in fact less than 10^^(gaggol+gaggol). One way to think about LAPL, and to have an intuitive feel for why its true, is to realize that (b{c}m){c}n is just (b{c}m){c-1){b{c}m){c-1} ... ... {c-1}(b{c}m) w/n (b{c}m)s. Relatively speaking having b{c}m in all of the terms makes little difference except in the case of the last one. Imagine replacing every b{c}m except the last with b. We thus obtain: b{c-1}b{c-1} ... {c-1}b{c-1}b{c}m This simplifies roughly to b{c}(m+n) which becomes the upperbound. This isn't rigorous enough to prove this is the upperbound because I first reduced the (b{c}m)'s to "b"s, but with a little more sophication it can be shown that the (b{c}m)s don't have nearly the effect initial intuition would suggest. In any case, try working out some problems and see if you can confirm LAPL for yourself. I'm fairly convinced the bounds are correct, but just in case some independent verification would be helpful in establishing it's validity. Sbiis Saibian (talk) 15:00, January 22, 2013 (UTC) I just stumbled on your unfinished article, when I tried to find more useful information on your site. Ikosarakt1 (talk) 17:25, January 22, 2013 (UTC) lol, seems you found one of my back doors. I'll have to fix that problem for future articles. Sbiis Saibian (talk) 04:18, January 24, 2013 (UTC) I concur with Sbiis; I too have known for some time that the lower up arrow notation grows about half as fast as the regular up arrow notation. So the Clarkkkkson is indeed greater than Graham's number and xkcd, since dividing by 2 is meaningless after the first iteration. Deedlit11 (talk) 15:17, February 28, 2013 (UTC) Wait, but in the function hypf(a,b,c) the parameter b responds to the number of down-arrows. But parameter d in the ck(a,b,c,d) iterates c in the hypf function, not the number of down-arrows. Ikosarakt1 (talk ^ 14:09, March 20, 2013 (UTC) Right, so ck(a,b,c,d) is just one level up the Ackermann hierarchy from hypf(a,b,c). So ck(n,n,n,n) does not outgrow Ackermann(n,n). It still grows about half as fast, so our assessment of the Clarkkkkson doesn't change. Deedlit11 (talk) 14:20, March 20, 2013 (UTC) Approximation? So if Clarkkkkson's recursion level is around \(\omega + 1\), does this mean that \(¥ \approx f_{\omega + 1}(K)\)? This is a problem I see in many articles on the wiki. We have a number defined as \(f(a)\), and another function \(g(n)\) has a growth rate similar to \(f(n)\). Does this really mean that \(g(a) \approx f(a)\) is even a reasonable assumption? FB100Z • talk • 06:49, April 6, 2013 (UTC) :It's not quite true in general that \(g(a) \approx f(a)\), for example Goodstein(n) \(= f_{\varepsilon_0}(\log_*(n) - 1)\). Still, it's "in the ballpark", as Goodstein \(>^* f_{\alpha}\) for all \(\alpha < \varepsilon_0\). So in general we can say that if \(g(n)\) has a growth rate similar to \(f(n)\) then their values will be in the same ballpark, googologically speaking. :In the case of Clarkkkkson, we indeed have \(¥ \approx f_{\omega + 1}(K)\). Deedlit11 (talk) 10:13, April 6, 2013 (UTC) Why did Clarkson use weak hyper-operators? I mean, they don't achieve even a growth rate of \(n\uparrow^3n\)! Can it be because of he didn't want so fast growth? 14:46, November 13, 2017 (UTC) :Read the above section carefully. It has been shown that the weak hyperoperators overtake \(f_k(n) = n\uparrow^kn\) for any fixed k. So using the weak hyperoperators doesn't make Clarkkkkson significantly smaller. -- ☁ I want more ⛅ 16:58, November 13, 2017 (UTC) ::But despite that it makes not much of a difference, it actually 'does''' make a (albeit very small) dent! 13:05, November 14, 2017 (UTC)